Question

Let p(x) be a polynomial of degree 3 and p(n) =1/n for n=1,2,3,4. Find the value of p(5)

Answer 1

We're given a polynomial p(x) of degree 3, such that, for $\sf{n=1,\ 2,\ 3,\ 4,}$

$\longrightarrow\sf{p(n)=\dfrac{1}{n}}$

$\longrightarrow\sf{n\cdot p(n)-1=0}$

Let $\sf{f(x)=x\cdot p(x)-1}$ such that $\sf{f(x)=0}$ for $\sf{x=1,\ 2,\ 3,\ 4.}$ This implies 1, 2, 3 and 4 are roots of $\sf{f(x).}$

$\Longrightarrow\sf{f(x)=k(x-1)(x-2)(x-3)(x-4)}$

for a constant $\sf{k\in\mathbb{R}.}$ Then by definition,

$\longrightarrow\sf{x\cdot p(x)-1=k(x-1)(x-2)(x-3)(x-4)\quad\quad\dots(1)}$

To find the value of $\sf{k,}$ let $\sf{x=0.}$ Then (1) becomes,

$\longrightarrow\sf{0\cdot p(0)-1=k(0-1)(0-2)(0-3)(0-4)}$

$\longrightarrow\sf{-1=k(-1)(-2)(-3)(-4)}$

$\longrightarrow\sf{-1=24k}$

$\longrightarrow\sf{k=-\dfrac{1}{24}}$

Hence (1) becomes,

$\longrightarrow\sf{x\cdot p(x)-1=-\dfrac{1}{24}\,(x-1)(x-2)(x-3)(x-4)}$

Let $\sf{x=5.}$ Then,

$\longrightarrow\sf{5\cdot p(5)-1=-\dfrac{1}{24}\,(5-1)(5-2)(5-3)(5-4)}$

$\longrightarrow\sf{5\cdot p(5)-1=-1}$

$\Longrightarrow\sf{\underline{\underline{p(5)=0}}}$

Answer 2

Answer:

We're given a polynomial p(x) of degree 3, such that, for \sf{n=1,\ 2,\ 3,\ 4,}n=1, 2, 3, 4,

\longrightarrow\sf{p(n)=\dfrac{1}{n}}⟶p(n)=

n

1

\longrightarrow\sf{n\cdot p(n)-1=0}⟶n⋅p(n)−1=0

Let \sf{f(x)=x\cdot p(x)-1}f(x)=x⋅p(x)−1 such that \sf{f(x)=0}f(x)=0 for \sf{x=1,\ 2,\ 3,\ 4.}x=1, 2, 3, 4. This implies 1, 2, 3 and 4 are roots of \sf{f(x).}f(x).

\Longrightarrow\sf{f(x)=k(x-1)(x-2)(x-3)(x-4)}⟹f(x)=k(x−1)(x−2)(x−3)(x−4)

for a constant \sf{k\in\mathbb{R}.}k∈R. Then by definition,

\longrightarrow\sf{x\cdot p(x)-1=k(x-1)(x-2)(x-3)(x-4)\quad\quad\dots(1)}⟶x⋅p(x)−1=k(x−1)(x−2)(x−3)(x−4)…(1)

To find the value of \sf{k,}k, let \sf{x=0.}x=0. Then (1) becomes,

\longrightarrow\sf{0\cdot p(0)-1=k(0-1)(0-2)(0-3)(0-4)}⟶0⋅p(0)−1=k(0−1)(0−2)(0−3)(0−4)

\longrightarrow\sf{-1=k(-1)(-2)(-3)(-4)}⟶−1=k(−1)(−2)(−3)(−4)

\longrightarrow\sf{-1=24k}⟶−1=24k

\longrightarrow\sf{k=-\dfrac{1}{24}}⟶k=−

24

1

Hence (1) becomes,

\longrightarrow\sf{x\cdot p(x)-1=-\dfrac{1}{24}\,(x-1)(x-2)(x-3)(x-4)}⟶x⋅p(x)−1=−

24

1

(x−1)(x−2)(x−3)(x−4)

Let \sf{x=5.}x=5. Then,

\longrightarrow\sf{5\cdot p(5)-1=-\dfrac{1}{24}\,(5-1)(5-2)(5-3)(5-4)}⟶5⋅p(5)−1=−

24

1

(5−1)(5−2)(5−3)(5−4)

\longrightarrow\sf{5\cdot p(5)-1=-1}⟶5⋅p(5)−1=−1

\Longrightarrow\sf{\underline{\underline{p(5)=0}}}⟹

p(5)=0